Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int maxn = 3e3+10;

double dp[maxn];

double p[maxn];

int main(void)
{
	int n;
	cin >> n;

	for (int i = 1; i <= n; i++)
		cin >> p[i];

	dp[0] = 1.00;
	for (int i = 1; i <= n; i++)
	{	
		for (int j = n; j >= 1; j--)
			dp[j] = (dp[j]*(1.00-p[i])+dp[j-1]*p[i]);

		dp[0] *= (1.00-p[i]);
	}

	double ans = 0.00;
	for (int i = (n+1)/2; i <= n; i++)
		ans += dp[i];

	printf("%.12lf\n", ans);
}